If ^ a _01 4+x^2 dx= 8 , find the value of a. - Vidyadip

Question

If $\int\limits^{\text{a}}_0\frac{1}{4+\text{x}^2}\text{ dx}=\frac{\pi}{8},$ find the value of a.

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Answer

$\int\limits^{\text{a}}_0\frac{1}{4+\text{x}^2}\text{ dx}=\frac{\pi}{8}$$\Rightarrow\frac{1}{2}\tan^{-1}\Big[\frac{\text{x}}{2}\Big]^{\text{a}}_0=\frac{\pi}{8}$ $\bigg[\int\frac{1}{\text{a}^2+\text{x}^2}\text{ dx}=\frac{1}{\text{a}}\tan^{-1}\frac{\text{x}}{\text{a}}+\text{C}\bigg]$
$\Rightarrow\frac{1}{2}\Big(\tan^{-1}\frac{\text{a}}{2}-\tan^{-1}0\Big)=\frac{\pi}{8}$
$\Rightarrow\tan^{-1}\frac{\text{a}}{2}-0=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\frac{\text{a}}{2}=\frac{\pi}{4}$
$\Rightarrow\frac{\text{a}}{2}=\tan\frac{\pi}{4}=1$
$\Rightarrow\text{a}=2$
Thus, the value of a is 2

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